Simplifying if(condition): return True to one line return statement changes results in Python

I do not understand why the two methods written below perform differently. Any insights into why the simplified statement in xyz_there() fails for this test case would be appreciated!

str = "abcxyz"

def xyz_there(str):
  for i in range(len(str)):
    return (str[i:i+3] == 'xyz' and str[i-1] != '.')
  return False

def xyz_there2(str):
  for i in range(len(str)):
    if str[i:i+3] == 'xyz' and str[i-1] != '.':
      return True
  return False

print(xyz_there(str))  
print(xyz_there2(str))

It has always been my understanding that the return(condition) was the most efficient way to write that expression. I am aware that functions like find() and count() can also be used to solve this problem.

Welcome to stackoverflow!

Difference between these two functions is that in second function, xyz_there2, the return True statement is inside if, so it is executed only if the condition is true. If it is not true - loop continues to next value of i.

However, in first function xyz_there, the return statement is not inside of any if statement, so it's executed on first iteration of the loop, thus returning False value. And after that the function exits, and loop doesn't get a chance to iterate to next value of i.

In other words, xyz_there function returns False at first loop iteration, when i==0. While xyz_there2 iterates until i==3, and returns True.

your first function is equivalent to this which is returning Flase :

def xyz_there(str):
  return (str[0:3] == 'xyz' and str[-1] != '.')

because you are not looping through all the characters in the string , you are just making decision on the first 3 letters of the word and return it